Find the Value of `(cos alpha+ cos beta)^2+(sinalpha+sin beta)^2`

Options

A. `4cos^2((alpha-beta)/2)`
B. `4cos^2((alpha+beta)/2)`
C. `4cos^2(alpha-beta)`
D. `cos^2((alpha+beta)/2)`

Explanation

To find the value of \((\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2\), we can follow these steps:

### Step 1: Expand the squares
We start by expanding both squares in the expression:

\[ (\cos \alpha + \cos \beta)^2 = \cos^2 \alpha + \cos^2 \beta + 2 \cos \alpha \cos \beta \]

\[ (\sin \alpha + \sin \beta)^2 = \sin^2 \alpha + \sin^2 \beta + 2 \sin \alpha \sin \beta \]

### Step 2: Combine the expanded expressions
Now, we combine the results from Step 1:

\[ (\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2 = (\cos^2 \alpha + \sin^2 \alpha) + (\cos^2 \beta + \sin^2 \beta) + 2 (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]

### Step 3: Use the Pythagorean identity
Using the Pythagorean identity, we know that:

\[ \cos^2 \alpha + \sin^2 \alpha = 1 \]

\[ \cos^2 \beta + \sin^2 \beta = 1 \]

So we can simplify the expression:

\[ = 1 + 1 + 2 (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \]

### Step 4: Use the cosine addition formula
The term \( \cos \alpha \cos \beta + \sin \alpha \sin \beta \) can be rewritten using the cosine of the difference of angles:

\[ \cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta) \]

### Step 5: Substitute back into the expression
Now, substituting this back into our expression gives:

\[ = 2 + 2 \cos(\alpha - \beta) \]

### Step 6: Factor the expression
We can factor out the 2:

\[ = 2(1 + \cos(\alpha - \beta)) \]

### Final Result
Thus, the value of \((\cos \alpha + \cos \beta)^2 + (\sin \alpha + \sin \beta)^2\) is:

\[ 2(1 + \cos(\alpha - \beta)) \]